Python模塊創建音樂播放列表（windows）

Question

我想創建一個簡單的python腳本，查看文件夾和子文件夾，並創建一個包含mp3文件夾名稱的播放列表。但到目前爲止，我只遇到了Python的模塊，在Linux上工作或我無法弄清楚如何安裝它們（pymad）..

這只是我的android手機所以想通過m3u格式應該做..我不關心任何其他元數據而不是mp3文件的名稱。

Answer 1

其實我只是看着http://en.wikipedia.org/wiki/M3U並認爲這是很容易寫的m3u文件...應該能夠簡單的Python寫做文本file`

這裏是我的解決方案

import os 
import glob 

dir = os.getcwd() 

for (path, subdirs, files) in os.walk(dir): 
    os.chdir(path) 
    if glob.glob("*.mp3") != []: 
     _m3u = open(os.path.split(path)[1] + ".m3u" , "w") 
     for song in glob.glob("*.mp3"): 
      _m3u.write(song + "\n") 
     _m3u.close() 

os.chdir(dir) # Not really needed.. 

Answer 2

我寫了一些代碼，會根據你的條件會返回所有嵌套播放列表的候選人名單：

import os 

#Input: A path to a folder 
#Output: List containing paths to all of the nested folders of path 
def getNestedFolderList(path): 

    rv = [path] 
    ls = os.listdir(path) 
    if not ls: 
     return rv 

    for item in ls: 
     itemPath = os.path.join(path,item) 
     if os.path.isdir(itemPath): 
      rv= rv+getNestedFolderList(itemPath) 

    return rv 

#Input: A path to a folder 
#Output: (folderName,path,mp3s) if the folder contains mp3s. Else None 
def getFolderPlaylist(path): 
    mp3s = [] 
    ls = os.listdir(path) 
    for item in ls: 
     if item.count('mp3'): 
      mp3s.append(item) 

    if len(mp3s) > 0: 
     folderName = os.path.basename(path) 
     return (folderName,path,mp3s) 
    else: 
     return None 

#Input: A path to a folder 
#Output: List of all candidate playlists 
def getFolderPlaylists(path): 
    rv = [] 
    nestedFolderList = getNestedFolderList(path) 
    for folderPath in nestedFolderList: 
     folderPlaylist = getFolderPlaylist(folderPath) 
     if folderPlaylist: 
      rv.append(folderPlaylist) 

    return rv 

print getFolderPlaylists('.')